Two blocks are sliding along a frictionless track. Block A (mass 4.03 kg) is moving to the right at 3.00 m/s. Block B (mass 4.84 kg) is movi

Question

Two blocks are sliding along a frictionless track. Block A (mass 4.03 kg) is moving to the right at 3.00 m/s. Block B (mass 4.84 kg) is moving to the left at 3.60 m/s. Assume the system to be both Block A and Block B. What is the total momentum of the system before the collision?

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Thanh Hà 3 weeks 2021-08-25T19:55:52+00:00 1 Answers 0 views 0

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    2021-08-25T19:57:44+00:00

    Answer:

    The total momentum of the system before the collision is 5.334 kg-m/s towards left.

    Explanation:

    Given that,

    Mass of the block A, m_A=4.03\ kg

    Speed of block A, v_A=3\ m/s

    Mass of the block B, m_B=4.8\ kg

    Mass of block B, u_B=-3.6\ m/s

    Let p is the total momentum of the system before the collision. It is given by :

    p=m_Av_A+m_Bv_B\\\\p=4.03\times 3+4.84\times (-3.6)\\\\p=-5.334\ kg-m/s

    So, the total momentum of the system before the collision is 5.334 kg-m/s towards left. Hence, this is the required solution.

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